INTERSECTION THEORY. Note that. If we identify the bidual V with V in a canonical fashion, then we deduce (U # ) # = U.

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1 INTERECTION THEORY LIVIU I. NICOLAECU ABTRACT. I provide more details to the intersection theoretic results in [1]. CONTENT 1. Transversality and tubular neighborhoods 1 2. The Poincaré dual of a submanifold 4 3. mooth cycles and their intersections 8 4. Applications The Euler class of an oriented rank two real vector bundle 18 References TRANVERALITY AND TUBULAR NEIGHBORHOOD We need to introduce a bit of microlocal terminology. We begin by reviewing a few facts of linear algebra. If V is a finite dimensional real vector space and U is a subspace of V, then its annihilator is the subspace U # := { v V ; v, u = 0, u U }. Note that dim U + dim U # = dim V = dim V. If we identify the bidual V with V in a canonical fashion, then we deduce If U 0, U 1 are subspaces in V then (U # ) # = U. (U 0 + U 1 ) # = U # 0 U # 1, (U 0 U 1 ) # = U # 0 + U # 1. (1.1) If T : V 0 V 1 is a linear map between finite dimensional vector spaces, then it has a dual It satisfies the equalities T : U 1 U 0. (ker T ) # = R(T ), R(T ) # = ker(t ) (1.2) where R denotes the range of a linear operator. Two linear maps T i : U 0 V, i = 0, 1 are said to be transversal, and we write this T 0 T 1, if R(T 0 ) + R(T 1 ) = V. Date: tarted Oct. 6, Completed on Oct. 14, Last revised October 31, Notes for Topics class, Fall

2 2 LIVIU I. NICOLAECU Using (1.1) and (1.2) we deduce that T 0 T 1 ker T 0 ker T 1 = 0, (1.3) i.e., T 0 T 1 if and only if the systems of linear equations T 0 v = 0, T 1 v = 0, v V, has only the trivial solution v = 0. uppose now that U 1 is a subspace in V and T 1 is the natural inclusion. In this case we say that T 0 is transversal to U 1 and we write this T 0 U 1. From (1.3) we deduce T 0 U 1 the restriction of T 0 to U # 1 Proposition 1.1. upposet that T 0 U 1. Then T 1 0 (U 1 ) # = T 0 (U # 1 ). is injective. (1.4) In particular, the codimension of T 1 0 (U 1 ) in U 0 is equal to the codimension of U 1 in V. Proof. Fix surjection T : V W such that ker T = U 1. Then Using (1.2) we deduce On the other hand T 1 0 (U 1 ) = ker T T 0. T 1 0 (U 1 ) # = R(T 0 T ) = T 0 (R(T )). R(T ) = (ker T ) # = U # 1. uppose now that is a smooth manifold of dimension m and is a closed submanifold of. The tangent bundle of is naturally a subbundle of the restriction to of the tangent bundle of, T (T ). The quotient bundle (T ) /T is called the normal bundle of in and it is denoted by T. In particular, we have a short exact sequence of vector bundles over Dualizing this sequence we obtain a short exact sequence 0 T (T ) T 0. (1.5) 0 (T ) (T ) T 0. The bundle (T ) is called the conormal bundle of in and it is denoted by T. It can be given the following intuitive description. uppose that (x 1,..., x m ) are local coordinates on an open set O such that in these coordinates the intersection O has the simple description x s+1 = = x m = 0, i.e., in these coordinates O can be identified with the coordinate plane (x 1,..., x s, 0,..., 0). We can regard the differentials dx s+1,..., dx m as defining a local trivialization of T on O. Note that for any x we have (T) x = (T x ) #. (1.6) Any Riemann metric on defines an isomorphism between the normal and conormal bundles of. From the exact sequence (1.5) we deduce that if two of the bundles T, (T ) T are orientable, then so is the third, and orientations on two induce orientations on the third via the base-first convention or(t ) = or T or T. (1.7)

3 w INTERECTION THEORY 3 A co-orientation on a closed submanifold is by definition an orientation on the normal bundle T. Thus, an orientation on the ambient space together with a co-orientation on induce an orientation on. Definition 1.2. uppose that is a closed submanifold of the smooth manifold. A tubular neighborhood of in is a pair (N, Ψ), where N is an open neighborhood of in, and Ψ : T N is a diffeomorphism such that its restriction to T N coincides with the identity map 1 :. We have the following fundamental result, [2, 3]. Theorem 1.3. (a) Let and be as in the above definition. Then for any open neighborhood U of in there exists a tubular neighborhood (N, Ψ) such that N U. (b) If (N 0, Ψ 0 ) and (N 1, Ψ 1 ) are two tubular neighborhoods of, then there exists a smooth map with the following properties. 1 (i) H 0 = 1. (ii) H t is a diffeomorphism of. (iii) H t (x) = x x, t [0, 1]. (iv) H 1 (N 0 ) = N 1. (v) The diagram below is commutative H : [0, 1], x H t (x), T [ Ψ 0 [[] Ψ 1 N 0 N 1 If (N, Ψ) is a tubular neighborhood of in, then we have a natural projection π N : N H 1 Ψ T π. We will denote by N x the fiber of π N over x. Observe that x N x. Remark 1.4. When is compact, we can construct a tubular neighborhood of as follows. Fix a Riemann metric g on. Then for any ε > 0 sufficiently small, the set N ε g := { x ; dist g (x, ) < ε } determines a tubular neighborhood of. The natural projection π N : N ε g has the following description: for x N ε g, the point π N (x) is the point on closest to x. If N ε g is as above, then the diffeomorphism Ψ : T N ε g can be constructed as follows. The normal bundle T can be identified with the orthogonal complement (T ) of T in (T ) and if x, X (T x ) T x, then Ψ(X) = exp g,x ( β( X g )X ), 1 A map H with the properties (i), (ii), (iii) above is called an isotopy (or diffeotopy) of rel.

4 4 LIVIU I. NICOLAECU where exp g,x : T x is the exponential map defined by the metric g, X g is the g-length of X, and β : [0, ) [0, ε) is a diffeomorphism such that β(t) = t if t < ε 4. uppose (respectively R, ) is a smooth manifold of dimension m (respectively r, s) and f : R, g : are smooth maps. We say that f is transversal to g, and we write this f g, if for any x R and y such that f(x) = g(y) = z the differentials f x : T x R T z and ġ : T y T z are transversal. When is a closed submanifold of and g is the natural inclusion, we say that f is transversal to and we write this f. Proposition 1.5. uppose is a manifold of dimension m, is a closed submanifold of of codimension c, R is a smooth manifold of dimension r and f : R is a smooth map transversal to. Then f 1 () is a closed submanifold of R of codimension c and there exists natural bundle isomorphism f : f T T f 1 () R, f : T f 1 () f T. Proof. The first claim is local so it suffices to prove it in the special case when coincides with a vector space of dimension V and is a subspace U of V of codimension c. Consider a surjection T : V R c such that ker T = U =. Fix x R, set y = f(x) and denote by f x the differential of f at x, f x : T x R T y = V. Observe that f 1 () = {z R; T (f(z)) = 0} The differential of T f at x R is T f x. ince f x (T x R) + T y = f x (T x R) + U 0 = T y = V and the linear map T is surjective with kernel U we deduce that T f x is surjective. We can now invoke the implict function theorem to conclude that that f 1 () is a submanifold. Using Proposition 1.1 and (1.6) we deduce that the dual of f x defines a linear isomorphism f x : (T ) f(x) (T 1 () R) x, x f 1 (). Observing that (T ) f(x) is the fiber at x f 1 () of the pullback f (T ) we deduce that these isomorphisms define a bundle isomorphism f : f (T ) T f 1 () R. The isomorphism f : T f 1 ()R f T is the dual of the above isomorphism. 2. THE POINCARÉ DUAL OF A UBANIFOLD Let be a finite type, oriented smooth manifold of dimension m. A (Borel-oore) cycle of dimension s in is defined to be a linear map map C : H s c () R, H s c () ω C, ω. The Poincaré dual of the cycle C is the cohomology class η C H m s () uniquely determined by the equality ω η C = C, ω, ω Hc s ().

5 INTERECTION THEORY 5 uppose that is a closed oriented submanifold of dimension s of the oriented manifold of dimension m. It defines an s-dimensional cycle [] via the equality [], ω = ω, ω Hc s (). The Poincaré dual of in is the Poincaré dual of the cycle [], i.e., cohomology class η = η H m s () uniquely determined by the equality ω η = ω, ω Hc s (). Theorem 2.1. uppose that π : E is an oriented real vector bundle of rank n over the oriented manifold. Denote by Φ E the Thom class of E and equip the total space of E using the base first convention. Denote by i : E the inclusion of as zero section. Then oreover, if is compact then E ω Φ E = η E = φ E. (2.1) ω, ω Ω s (E), dω = 0. (2.2) We want to emphasize that in the above equality we do not require that ω have compact support. Proof. Define H : E [0, 1] E, E [0, 1] (z, t) H t (z) := (1 t) z + t i π(z), where t denotes the multiplication by t along the fibers of E. Observe that H is proper, H 0 = 1 E, H 1 = i π. Thus for any closed, compactly supported ω Ω s c(e) there exists a τ Ω s 1 c (E) such that ω π i ω = H0 ω H1 ω = dτ. We have (use the projection formula) E ω Φ E = = E π (i ω) Φ E + d(τ Φ E ) E } {{ } =0 i ω π Φ E = This proves (2.1). To prove (2.2) use the same argument and the fact that when is compact, the Thom class has compact support. Corollary 2.2. uppose that is a compact s-dimensional oriented submanifold of the smooth, finite type oriented manifold of dimension m. Denote by Φ T N the Thom class of the normal bundle with the orientation induced via the base-first convention (1.7). Fix a tubular neighborhood (N, Ψ) of in, denote by η N the Poincaré dual of in N, and by j the extension by 0 morphsim Hc (N) H (). Then i ω. η N = (Ψ 1 ) Φ T, η = j η N. (2.3)

6 6 LIVIU I. NICOLAECU Remark 2.3. Let us point out a rather subtle phenomenon. uppose that is a compact s-dimensional oriented submanifold of the oriented manifold of dimmension m. uppose further that N is a tubular neighborhood of in Observe that defines defines a linear map H s () ω ω R. From the Poincaré duality we deduce that there exists a unique compactly supported cohomology class η comp Hc m s () such that ω η comp = ω, ω H s (). We have a natural morphism j # : Hc m s () H m s () and the above corollary proves that η = j # (η comp ). In particular, we deduce that two closed compactly supported forms η 0, η 1 representing η must be cohomologous as compactly supported forms, i.e., there exists a compactly supported form α such that η 1 η 0 = dα. Corollary 2.4. uppose that is a compact s-dimensional oriented submanifold of the smooth finite type oriented manifold of dimension m. A cohomology class u H m s () is the Poincaré dual of if and only if there exists a tubular neighborhood (N, Ψ) of in such that u is represented by a closed form η with compact support contained in N and such that N x η = 1, x, where N x is oriented as the fiber of (T ) x via the diffeomorphism Ψ : (T ) x N x. Definition 2.5. Let, and L be closed, oriented submanifolds of the oriented manifold N such that dim + dim L = dim N (a) If p L is a point where the two submanifolds intersect transversally, i.e., T p L+T p = T p N, then we define ɛ(p) = ɛ(, L, p) {±1} via the rule or(t p ) or(t p L) = ɛ(, L, p) or(t p N). We will refer to or(l, p) as the local intersection number of L and at p. Note that or(l,, p) = ( 1) dim L dim or(, L, p). (b) If is compact, and L, then L is finite, and we define the intersection number of the cycles L and to be the integer [] [L] := ɛ(, L, p). p L Theorem 2.6. uppose that, L are closed oriented submanifolds of the smooth, finite type, oriented manifold N satisfying the following conditions. is compact. dim L + dim = dim N. L.

7 INTERECTION THEORY 7 If η is compactly supported closed form representing the Poincaré dual of., and η L is a closed form representing the Poincaré dual of [L],Then [] [L] = η L η = η. (2.4) Proof. et s := dim, l := dim L, n = s + l = dim N. For simplicity, for any p L we will set ɛ(p) = ɛ(, L, p). From the transversality of the intersection we deduce that for any p L we can find local coordinates (x 1,..., x n ) defined on an open neighborhood U p of p such that U p = { x s+1 = = x s+l = 0 }, U p L = { x 1 = = x s = 0 }, the orientation of U is described by the volume form dx 1 dx s, the orientation of L U is given by dx s+1 dx s+l. Under these assumptions we deduce that L or(u p ) = ɛ(p)dx 1 dx n. (2.5) We can additionally assume that U p U q = if p q. We can find neighborhoods V p of p in U p and a Riemann metric g on N such that along V p the metric g has the Euclidean form g = (dx 1 ) (dx n ) 2. L V p p N FIGURE 1. Two manifolds of complementary dimensions intersecting transversally Construct a thin tubular neighborhood (N, Ψ) of in N such that for every p L we have (see Figure 1) N p = (L V p ) N =: L p where we recall that N p denotes the fiber of N over p with respect to the natural projection π N : N. We can choose N thin enough so that we have the aditional equality L N = N p. p L Consider a compactly supported closed form η N Ωl c(n) representing the Poincaré dual of in N. The fiber N p is equipped with a natural orientation, given by the co-orientation of in N. On the other hand L p is equipped with an orientation induced by the orientation of L. The equality (2.5) implies or(l p ) = ɛ(p) or(n p ).

8 8 LIVIU I. NICOLAECU Now observe that η η L = L η = p L L p η = p L ɛ(p) η = N p p L ɛ(p). 3. OOTH CYCLE AND THEIR INTERECTION uppose that is a smooth, compact oriented manifold of dimension m, R is a smooth compact oriented manifold of dimension r and f : R is a smooth map. The pair (R, f) defines a linear map H r () R, H r () ω f ω. We will denote this element of H r () by f [R] and we will refer to it as the smooth cycle of dimension r determined by the pair (R, f). From the Poincaré duality we deduce that there exists η f [R] H m r () such that f ω = ω η f [R], ω H r (). R We will refer to the cohomology class η f [R] as the Poincaré dual of the smooth cycle f [R]. uppose that R 0, R 1 are two compact oriented manifolds of the same dimension r and f i : R i are smooth maps, i = 0, 1. We say that the pairs (R 0, f), (R 1, f 1 ) are cobordant, and we write this (R 0, f 0 ) c (R 1, f 1 ) if there exists a compact, oriented (r + 1)-dimensional manifold with boundary R and a smooth map f : R such that the following hold. R = R 0 R 1. The orientation induced by R 1 coincides with the orientation of R 1, while the orientation induced by R on R 0 is the opposite orientation of R 0. f Ri = f i, i = 0, 1. The pair ( R, f) is called an oriented cobordism between (R 0, f 0 ) and (R 1, f 1 ). Proposition 3.1. If (R 0, f 0 ) (R 1, f 1 ) then (f 0 ) [R 0 ] = (f 1 ) [R 1 ]. R Proof. We have to show that for any ω Hc r (), r = dim R 0 = dim R 1 we have f0 ω = f1 ω. R 0 R 1 Let ( R, f) be an oriented cobordism between (R 0, f 0 ) and (R 1, f 1 ) and ω Hc r () then f1 ω f0 ω = F ω = d f ω R 1 R 0 R b br = f (dω) = 0. br

9 INTERECTION THEORY 9 uppose that is a smooth, compact, oriented submanifold of of codimension k. Its conormal bundle T is equipped with a natural orientation given by the rule or(t ) = or(t ) = or() or(t). Denote by i the canonical inclusion. Observe that η = η i [] H k (). Two compact, oriented submanifolds i 0 : 0, (i 1 : 1 of the same dimension s are said to be cobordant if the pairs ( 0, i 0 ) and ( 1, i 1 ) are cobordant. We write this 0 c 1. Proposition 3.1 shows that 0 c 1 η 0 = η 1. Corollary 3.2. uppose that 0, 1 are two compact, oriented s-dimensional manifolds of the oriented, finite type smooth manifold of dimension m, and L is a closed, oriented submanifold of of dimension m s that intersects transversally both 0 and 1. If 0 c 1, then [ 0 ] [L] = [ 1 [L]. Proof. ince 0 c 1 we have η 0 = η 1. The corollary now follows form Theorem 2.6. Example 3.3. For t [0, 1] define F t : CP 1 CP 2, CP 1 [z 0, z 1 ] [tz 0, (1 t)z 1, z 1 ] CP 2, where [z 0, z 1,..., z n ] denote the homogeneous coordinates in CP n. Observe that each F t is an embedding. The image of F 0 is the projective line while the image of F 1 is the projective line H 0 = { [ζ 0, ζ 1, ζ 2 ]; ζ 0 = 0 }, H 1 = { [ζ 0, ζ 1, ζ 2 ] CP 2 ; ζ 1 = 0 }. We deduce that η H0 = η H1 =: η H. Observe that H 0 H 1 and the unique intersection point is p 0 [0, 0, 1]. An lementary computation shows that ɛ(p 0 ) = 1. Hence η H η H = [H 0 ] [H 1 ] = 1. CP 2 uppose that R is a compact oriented manifold of dimension r, is a compact oriented submanifold of codimension k, and f : R is a smooth map transversal to. Using Proposition 1.5 we deduce that the submanifold f 1 () of R carries a co-orientation induced by the bundle isomorphism f : f (T) Tf 1 () R. We thus have an orientation on f 1 (). We obtain a smooth (r k)-dimensional cycle [] f [R] := f [f 1 ()]. Denote by η [] f [R] the Poincaré dual of this cycle, i.e., ω η [] f [R] = f ω, ω H r k (). (3.1) f 1 ()

10 10 LIVIU I. NICOLAECU Theorem 3.4. Let, R, and f as above. Then f η = ηf R 1 (). (3.2) η [] f [R] = η η f [R]. (3.3) Proof. Let first observe that (3.3) is an immediate consequence of (3.2). Indeed, for any ω H r k () we have ω η η f [R] = f (ω η ) = f ω f η (3.2) = R f ω η R f 1 () = R f 1 () f ω (3.1) = R ω η [] f [R]. To prove (3.2) we will rely on Corollary 2.4. Fix a Riemann metric h on and a metric g on R. For simplicity we set X = f 1 () R. Form the tubular neighborhood Nh ε of in and Nρ g of X in R as explained in Remark 1.4. For x X we denote by Ng(x) ρ the fiber over x of the natural projection Ng ρ X and by Nh ε(y), y = f(x) the fiber of the natural projection Nε h. Observe that Ng(x) ρ = { ( ) z R; dist g z, X = distg (z, x) < ρ }. Lemma 3.5. There exists ρ 0 > 0 sufficiently small such that the restriction of f to N ρ 0 g (x) is an embedding for any x X. Proof. Denote by f x the differential of f at x. It is a linear operator f x : T x R T y, whose kernel is T x X. Thus, its restriction to T x Ng(x) ρ = (T x X) is injective. We can then find a neighborhood U x of x and ρ(x) > 0 such that for any x U the restriction of f to Ng ρ(x) (x ) is an embedding. Now cover X with finitely many neighborhood U x1,..., U xν and set ρ 0 = min { ρ(x 1 ),..., ρ(x ν ) }. For ρ < ρ 0 we set L ρ x := f ( N ε g(x) ). R X N x ρ g (x) f ρ x L f(x) FIGURE 2. apping the normal slice N ρ g(x). Lemma 3.6. Fix ρ < ρ 0. There exists ε = ε(ρ) > 0 such that for any x X the intersection L ρ x N ε h is closed in N ε h.

11 Proof. et INTERECTION THEORY 11 Σ ρ := { z R; dist g ( z, X ) = ρ }. The set f(σ ρ ) is compact and disjoint from and thus d(ρ) := dist h ( f(σρ ), ) > 0. Let ε = 1 2 d(ρ). We claim that that Lρ x N ε h is closed in Nε h. uppose that (y n) is a sequence in L ρ x N ε h that converges to a point y N ε h. We need to show that y L ρ x. By construction, there exist points p n N ρ g(x) such that f(p n ) = x n. The sequence (p n ) has a subsequence that converges to a point p N g (x) such that dist (p, x) r. Then f(p ) = y and dist(y, ) < ε < d(ρ). This implies that p Σ ρ because dist(f(σ ρ ), ) > d(ρ). Hence dist(p, x) = dist(p, ) < ρ so that p Ng(x) ρ and thus y L ρ x. Let ε = ε(ρ) as in the above Lemma. Denote by η ε a closed form representing the Poincaré dual of having support in Nh ε. According to Corollary 2.4 the differential form f η ε represents the Poincaré dual of X = f 1 () in R if We have N ρ g(x) f η ε Ng(x) ρ = 1, x X. f η ε = L ρ X η ε = L ρ x N ε h where L ρ x is equipped with the orientation induced by the diffeomorphism f : Ng(x) ρ L ρ x. The orientation on the normal slice Ng(x) ρ is the co-orientation of X which is determined by the coorientation of. Hence the tangent space of L ρ x at x is equipped with the coorientation of at x. Our choice of ɛ guarantees that L ρ x Nh ε is closed in Nε h. Theorem 2.6 now implies that η ε = [] [L ρ x] = 1. L ρ x N ε h In the special case when f : R is an embedding, the equality (3.2) can be reformulated as follows. Corollary 3.7. uppose that R, are compact oriented submanifolds of the oriented finite type manifold and R. Then R is a smooth submanifold of R. We orient R using the natural isomorphism T R R = (T ) R. Then the Poincaré dual of R in R is equal to the restriction to R of the Poincaré dual of in. η ε, We want to define a new operation on smooth cycles. uppose that 0, 1 are smooth oriented, finite type manifolds of dimensions m 0 and respectively m 1. Let f i : R i i, i = 0, 1 be smooth maps, where R i is a smooth compact manifold of dimension r i. These maps define smooth cycles (f i ) [R i ], i = 0, 1, with Poincaré duals η (fi ) [R i ] H m i r i ( i ).

12 12 LIVIU I. NICOLAECU We have a natural bilinear map : H ( 0 ) H ( 1 ) H ( 0 1 ), α 0 α 1 = (π 0α 0 ) (π 1α 1 ), where π i : 0 1 i, i = 0, 1, denotes the canonical projection. We have a smooth map f 0 f 1 : R 0 R Equip R 0 R 1 and 0 1 with the product orientations. We obtain a smooth cycle with Poincaré dual Proposition 3.8. (f 0 f 1 ) [R 0 R 1 ] η (f0 f 1 ) [R 0 R 1 ] H (m 0+m 1 ) (r 0 +r 1 ) ( 0 1 ). η (f0 f 1 ) [R 0 R 1 ] = ( 1) r 1(m 0 r 0 ) η (f0 ) [R 0 ] η (f1 ) [R 1 ]. (3.4) Proof. We have α 0 α 1 η (f0 f 1 ) [R 0 R 1 ] = (f 0 f 1 ) (α 0 α 1 ) = 0 1 R 0 R 1 ( ) ( ) f0 α 0 f1 α 1 R 0 R 1 = f0 α 0 R 0 f0 α 1 R 1. On the other hand (α 0 α 1 ) (η (f0 ) [R 0 ] η (f1 ) [R 1 ]) = 0 1 π0α 0 π1α 1 π0η (f0 ) [R 0 ] π1η (f1 ) [R 1 ] 0 1 = ( 1) r 1(m 0 r 0 ) π0(α 0 η (f0 ) [R 0 ]) π1(α 1 η (f1 ) [R 1 ]) 0 ( 1 ) ( ) = ( 1) r 1(m 0 r 0 ) α 0 η (f0 ) [R 0 ] α 0 η (f1 ) [R 1 ] 0 ( ) ( 1 ) = ( 1) r 1(m 0 r 0 ) f0 α 0 R 0 f0 α 1 R 1. Künneth theorem implies that any cohomology class in 0 1 is a linear combination of cohomology classes of the form α 0 α 1. The equality (3.4) is now obvious. Example 3.9. uppose that is a smooth compact oriented manifold of dimension and f : is a smooth map. Then the graph of f is a smooth submanifold Γ f equipped with a natural orientation induced by the diffeomorphism δ f : Γ f, x (x, f(x)). We want to offer a description of of the Poincaré dual η Γf H m ( ). Fix a basis (ω j ) 1 i b() of H (), b() = k b k(), where ω i denotes a closed from of degree ω i. Denote by (ω j ) 1 j N the Poincaré dual basis of H (), i.e., the forms ω j satisfy the equalities 2 { ω i ω j = δj i 1, i = j := 0, i j. Above ω j is a form of degree ω j = n ω j. 2 Throughout we use the convention that R α = 0 if α is a differential form such that deg α m.

13 INTERECTION THEORY 13 Kunneth s theorem then implies that the collection Ω i j := ωi ω j H ( ), 1 i, j b() is a basis of H (). imilarly, the collection Θ i j = ω j ω i is a basis of H ( ) and therefore, Poincaré dual of Γ f has the form η Γf = i,j c j i Θi j. There exist real numbers (a i j ) such that f ω j = i a i jω i, j. ore precisely a i j = ω i f ω j. We want to express the coefficients c i j in terms of the coefficients ai j. To achieve this we use the defining property of η Γf, i.e. Ω k l η Γ f = Ω k l = δf Ωk l. Γ f Observe first that δf Ωk l = ω k f ω l = j a j l ω k ω j = j a j l δk j = a k l. For i = 0, 1 denote by π i : the projection (x 0, x 1 ) x i. We have Ω k l η Γ f = c j i Ω k l Θi j i,j = c j i (π0ω k π1ω l ) (π0ω j π1ω i ) i,j = ( 1) ω l ω j c j i π0(ω k ω j ) π1(ω l ω i ) i,j = ( ) ( ) ( 1) ω l ( ω j + ω i ) c j i ω k ω j ω i ω l i,j = ( 1) ω l ( ω j + ω i ) c j i δk j δl i = ( 1) ω l ( ω k + ω l ) c k l. i,j Hence c k l := ( 1) ω l ( ω k + ω l ) a k l = ( 1) ω l ( ω k + ω l ) ω k f ω l. (3.5)

14 14 LIVIU I. NICOLAECU 4. APPLICATION uppose that N is a compact oriented smooth manifold of even dimension 2m. For any compact oriented submanifold of middle dimension m we define the self intersection number [] [] to be [] [] := η η. One can show (see [2, Thm. 14.6]) then there exists a smooth map F : [0, 1], (t, x) F t (x) such that for any t the map F t is an embedding, F 0 coincides with the inclusion i : and for any t > 0 the map F t is transversal to. If we denote by t the image of F t then we deduce that t, η t = η and thus [] [] = [ t ] [] = ɛ( t,, p) t (0, 1]. p t The computations in Example 3.3 can be reformulated as [H] [H] = 1. Theorem 4.1 (Lefschetz fixed point theorem). uppose is a compact oriented smooth manifold of dimension m and f : is a smooth map. We define its Lefschetz number to be the quantity m ( L(f) = ( 1) k tr k f, tr k f := tr f : H k () H k () ) Then Hence, if Γ f then k=0 η Γf η = L(f). [Γ f ] [ ] = L(f). In particular, we deduce that if L(f) 0, then the map f must have at least one fixed point. Proof. We continue to use the notations in Example 3.9. We have η Γf = i,j c j i ω j ω i. Then η Γf η = η Γf = i,j c j i ω j ω i = i,j c j i ω j ω i = i,j ( 1) ωi ω i c j i ω i ω j = i,j ( 1) ωi ω j c j i δi j = i ( 1) ωi ω i c i i (3.5) = i ( 1) ωi ω i +m ω i a i i = i ( 1) ω i a i i = k ( 1) k a i i. ω i =k }{{} =tr k f

15 INTERECTION THEORY 15 Corollary 4.2. uppose that is a compact oriented manifold of dimension m and denotes the diagonal in, i.e., the submanifold = { (x, x); x }. It is tautologically diffeomorphic to and thus it carries a natural orientation. Then [ ] [ ] = χ() = the Euler characteristic of. (4.1) Proof. This follows from the Lefschetz fixed point theorem by observing that is the graph of the identity map 1 and that tr k (1 ) = b k (). Definition 4.3. uppose that E is a smooth, oriented vector bundle of rank r over the compact smooth manifold. The Euler class of E is the cohomology class e(e) H r () defined by the equality e(e) = i Φ E, where Φ E Hc r (E) denotes the Thom class of E and i : E denotes the inclusion of into E as zero section. Theorem 4.4. Let π : E be as above. 3 If E admits a nowhere vanishing section s : E, then e(e) = 0. Proof. Fix a metric g on E. For any ε > 0 we denote by D ε (E) the open subset consisting of the vectors of length < ε in all the fibers of E. Fix ε > 0 such that s(x) g > ε, x. Next choose a representative Φ ε Ω r c(e) of Φ E such that supp Φ ε D ε (E). By construction, the image of s and the support of Φ ε are disjoint. Hence s Φ ε = 0. On the other hand, the form s Φ ε represents the cohomology class e(e). Indeed consider H : [0, 1] E, (t, x) t s(x), where t denotes the multiplication by t along the fibers of E. Observe that H 0 = i and H 1 = s. Hence the maps i and s are homotopic and therefore the forms i Φ ε and s Φ ε are cohomologous. Remark 4.5. The above theorem implies that if E is a trivial vector bundle, then e(e) = 0 because trivial vector bundles admit nowhere vanishing sections. We can turn this result on its head and conclude that if e(e) 0, then the bundle E cannot be trivial. In other words, the Euler class can be interpreted as a measure on nontriviality of E. Theorem 4.6. uppose that E is a smooth, oriented real vector bundle of rank r over the smooth, compact, oriented manifold of dimension m r. uppose that s : E is a smooth section that is transversal to the zero section i : E. Denote by Z the zero set of s. Then Z is a smooth submanifold of of codimension r and we have a natural bundle isomorphism T Z = E. 3 Observe that we impose no orientability assumption on.

16 16 LIVIU I. NICOLAECU The above isomorphism equips Z with a coorientation and orientation and the Poincaré dual of the cycle [Z] is equal to the Euler class e(e). Proof. This is a special case of Corollary 3.7 applied to the submanifolds R = i() and = s() of the total space of E. Theorem 4.7. uppose that E is a smooth, oriented real vector bundle of rank m over the smooth, compact, oriented manifold of dimension m. uppose that s : E is a smooth section that is transversal to the zero section i : E. Then [] [s()] = e(e). (4.2) Proof. Denote by η The Poincaré dual of in E and by η s(] the Poincaré dual of s() in E. Then η = Φ E = the Thom class of E. We have [] [s()] = η η s [] = Φ E η s [] = s Φ E. E As in the proof of Theorem 4.4 we observe that s Φ E = i Φ E = e(e) in H m () since the sections s and i are homotopic. Remark 4.8. Observe that the intersection s() = i() s() is the set of zeros of the section s. The above theorem then states that a certain signed count of zeros of a transversal section is equal to the integral of the Euler class over the base when the dimension of the base is equal to the rank of the bundle. The sign associated to a transverse zero can be computed as follows. uppose that U is a coordinate patch diffeomorphci to R n with local coordinates x 1,..., x m such that the orientation of along U is given by dx 1 dx m. Fix an orientation preserving trivialization E U R m U. we get m-function y 1,..., y m : E U R, describing linear coordinates long fibers such that the orientation of each fiber is given by dy 1 dy m. A section of s is described over U as a map U R m, y j = y j (x 1,..., x m ), j = 1,..., m. If p 0 = (x 1 0,..., xm 0 ) is a zero of s in U, then the intersection of s() with at p 0 is transversal if and only if det (y1,..., y m ) (x 1,..., x m ) p 0 0. If this is the case, then zero p 0 is counted with sign ɛ(p 0 ) = sign det (y1,..., y m ) (x 1,..., x m ) p 0. One can show that this sign is independent of the various choices. Definition 4.9. Let be a compact oriented smooth manifold of dimension m. Then the Euler class of is the cohomology class e() := e(t ) H m (). E Theorem 4.10 (Poincaré-Hopf). e() = χ().

17 INTERECTION THEORY 17 Proof. Consider the diagonal and its normal bundle E := T. We identify E with a tubular neighborhood N of in and the Thom class Φ E of E with the Poincaré dual η of in N, and by extension, with the Poincaré dual of in. We have e(e) = Φ E = N η η (4.1) = χ(). The diagonal is clearly isomorphic to. To complete the proof the theorem wee need the following result. Lemma The normal bundle T ( ) is orientation preservingly isomorphic to T = T. Proof of the Lemma. Note that we have a short exact sequence of bundles over =, 0 T δ T T where for any x the map δ x : T x T x T x is given by v v v, while the map p x : T x T x T x p T 0 is given by p x (v 0, v 1 ) = v 1 v 0 2. The image of δ coincides with the tangent bundle T and thus p induces an isomorphism T ( ) T. This isomorphism is orientation preserving. To see this observe first that the above short exact sequence sequence admits a splitting σ : T T T, p σ = 1 T, given by σ x (v) = v v, x, v T x. To check that the isomorphism p : T ( ) T is orientation preserving it suffices to check that for any x we have Fix an oriented basis e 1,..., e m of T x. et or δ(t x ) or σ(t x ) = or(t x T x ). u i := e i 0 T x T x, v j = 0 e j T x T x. The collection u 1,..., u m, v 1,..., v m is an oriented basis of T x T x. Observe that Then We have δ(e i ) = u i + v i, σ(e j ) = v j u j. or δ(t x ) or σ(t x ) = or ( δ(e 1 ) δ(e m ) σ(e i ) σ(e m ) ). δ(e i ) δ(e m ) σ(e i ) σ(e m ) = ( 1) m(m 1) 2 δ(e 1 ) σ(e 1 ) δ(e m ) σ(e m ). Now we observe that and we conclude δ(e j ) σ(e j ) = 2u j v j ( 1) m(m 1) 2 δ(e i ) δ(e m ) σ(e i ) σ(e m ) = u 1 v 1 u m v m = ( 1) m(m 1) 2 2 m u 1 u m v 1 v m. This completes the proof of the lemma and thus of Theorem 4.10.

18 18 LIVIU I. NICOLAECU 5. THE EULER CLA OF AN ORIENTED RANK TWO REAL VECTOR BUNDLE uppose that π : E is an oriented rank two real vector bundle. We denote by Φ E H 2 cv(e) its Thom class and by e(e) H 2 () its Euler class defined by e(e) = i Φ E, where i : E denotes the inclusion of in E as zero section. We want to give an explicit description of Φ E and e(e). To this aim we fix a metric on E and a good (coordinate cover) {U α } α A that trivializes E. Thus E is defined by a gluing cocycle ore precisely, we can write g βα (x) = g βα : U αβ O(2). where ϕ βα : U αβ R is a smooth function. oreover The cocycle condition is then equivalent to the condition [ ] cos ϕβα sin ϕ βα, sin ϕ βα cos ϕ βα ϕ βα = ϕ αβ,. (5.1) g αγ (x) g γβ (x) g βα (x) = 1, x U αβγ ϕ αγ (x) + ϕ γβ (x) + ϕ β (x) 2πZ, x U αβγ. ince U αβγ is connected we deduce that for any α, β, γ there exists n γβα Z such that ϕ αγ (x) + ϕ γβ (x) + ϕ βα (x) = 2πn γβα, x U αβγ. (5.2) the correspondence (α, β, γ) n γβα is skew-symmetric in its arguments, i.e., n γβα = n βγα = n γαβ = n αβγ. The metric and the orientation on E induce polar coordinates (r α, θ α ) on E 0 Uα where E 0 denotes complement of the zero section, E \ i(). On the overlap E 0 Uαβ we have r α = r β, but the angular coordinates θ β and θ α are related by the equalities Lemma 5.1. There exist ξ α Ω 1 (U α ) such that for any α, β we have θ β θ α = π ϕ βα. (5.3) 1 2π dϕ βα = ξ β ξ α on U αβ. (5.4) Proof. Consider a partition of unity (ρ γ ) subordinated to (U γ ). Now define ξ α = 1 ρ γ dϕ γα. 2π Then (5.1) = 1 2π γ ξ β ξ α = 1 2π ρ γ d(ϕ αγ + ϕ γβ ) (5.2) = 1 2π γ ρ γ d(ϕ γα ϕ γβ ) γ γ ρ γ d(ϕ βα 2πn γβα ) = 1 2π dϕ βα.

19 From (5.4) we deduce that INTERECTION THEORY 19 dξ α = dξ β on U αβ. Thus the collection of closed 2-forms (ω α = dξ α ) defines a closed form ω Ω 2 (). Proposition 5.2. The cohomology class determined by ω is the Euler class e(e). Proof. The equalities (5.3) and (5.4) show that for any α, β we have the equality 1 2π dθ α π ξ α = 1 2π dθ β π ξ β, on E 0 Uαβ. We deduce that the collection of 1-forms ψ α := 1 2π dθ α π ξ α determines a global 1-form ψ Ω 1 (E 0 ) with the property that for any U α and any x U α the restriction to the fiber Ex 0 coincides with the angular form 1 2π dθ α. For this reason we say that ψ is a global angular form. The form ψ is defined outside the zero section. Note that dψ α = π dξ α = π ω α so that dψ = π ω, so that dψ extends to a form on E. Consider a smooth function ρ : [0, ) [ 1, 0] whose graph has the shape depicted in Figure 3-1 FIGURE 3. A smooth cut-off function ρ(r). Define Φ = d ( ρ(r)ψ ) Ω 2 (E 0 ), where r : E [0, ) denotes the radial distance function defined by the metric on E. We have Φ = dρ(r) ψ ρ(r)π ω. (5.5) The form dρ is identically zero near the zero section and we deduce that Φ extends as a closed 2-form on E. From the shape of ρ we deduce that it has compact vertical support. If x then Φ = dρ(r) ψ. E x E x On E x we have ψ = 1 2π dθ= the angular form on E x. We deduce Φ = 1 dρ(r) dθ, E x 2π R 2 where (r, θ) denote the angular coordinates on R 2. We have 1 1 dρ(r) dθ = lim 2π R 2 ε 0 2π r ε dρ(r) dθ

20 20 LIVIU I. NICOLAECU (use tokes formula) 1 = lim ρ(ε)dθ = 1 ε 0 2π r=ε since ρ(ε) = 1 if ε is sufficiently small. Thus Φ represents the Thom class of E. We deduce i Φ = ρ(0)ω = ω. The conclusion follows from the definition of the Euler class as the pullback of the Thom class via the zero section. REFERENCE [1] R.Bott, L. Tu: Differential Forms in Algebraic Topology, pringer Verlag, [2] Th. Bröcker, K. Jänich: Introduction to Differential Topology, Cambridge University Press, 2002 (2007). [3].W. Hirsch: Differential Topology, pringer Verlag, DEPARTENT OF ATHEATIC, UNIVERITY OF NOTRE DAE, NOTRE DAE, IN address: nicolaescu.1@nd.edu URL: lnicolae/